To start off, I am a beginner in python so I am not even sure if my question makes sense or is even possible.
I have 2 python files app.py. and compare.py. compare.py takes in two arguments (File paths) to run. So for example, when I want to run it, I do python compare.py ./image1.jpg ./image2.jpg. Now the return I get is some text printed to the terminal such as Comparison Done, The distance is 0.544.
Now, I want to run this compare.py from inside app.py and get a string with whatever compare.py would usually output to the terminal. So for example:
result = function('compare.py ./image1.jpg ./image2.jpg') and result will have the required string. Is this possible?
You can use os.popen:
In app.py:
import os
output = os.popen('python compare.py ./image1.jpg ./image2.jpg').readlines()
Related
I'll want to know how to call a function in vs code. I read the answer to similar questions, but they don't work:
def userInput(n):
return n*n
userInput(5)
And appends nothing
def Input(n):
return n*n
And in the terminal:
from file import *
from: can't read /var/mail/file
Can somebody help me?
You are doing everything correctly in the first picture. In order to call a function in python on vs code you first have to define the function, which you did by typing def userInput(n):. If you want to see the result of your function, you should not use return, you should use print instead. Return is a keyword- so when your computer reaches the return keyword it attempts to send that value from one point in your code to another. If you want to see the result of your code, typing print (n) would work better.
Your code should look like this:
def userInput(n):
print (n * n)
userInput(5)
The code would print the result 25
Your terminal is your general way to access your operating system, so you have to tell it that you want it to interpret your Python code first.
If you want to run the file you're typing in, you have to first know the location of that file. When you type ls in your terminal, does the name of your Python file show up? If not, hover over the tab in VSCode (it's close to the top of the editor) and see what path appears. Then in your terminal type cd (short for "change directory") and then the path that you saw, minus the <your filename here>.py bit. Type ls again, and you should see your Python file. Now you can type python <your filename here>.py to run it (provided you have Python installed).
You could also run the IDLE by just typing python in your terminal. This will allow you to write your code line-by-line and immediately evaluate it, but it's easier to write in VSCode and then run it with the method I described before.
I have written multiple python scripts that are to be run sequentially to achieve a goal. i.e:
my-directory/
a1.py,
xyz.py,
abc.py,
....,
an.py
All these scripts are in the same directory and now I want to write a single script that can run all these .py scripts in a sequence. To achieve this goal, I want to write a single python(.py) script but don't know how to write it. I have windows10 so the bash script method isn't applicable.
What's the best possible way to write an efficient migration script in windows?
using a master python script is a possibility (and it's cross platform, as opposed to batch or shell). Scan the directory and open each file, execute it.
import glob,os
os.chdir(directory) # locate ourselves in the directory
for script in sorted(glob.glob("*.py")):
with open(script) as f:
contents = f.read()
exec(contents)
(There was a execfile method in python 2 but it's gone, in python 3 we have to read file contents and pass it to exec, which also works in python 2)
In that example, order is determined by the script name. To fix a different order, use an explicit list of python scripts instead:
for script in ["a.py","z.py"]:
That method doesn't create subprocesses. It just runs the scripts as if they were concatenated together (which can be an issue if some files aren't closed and used by following scripts). Also, if an exception occurs, it stops the whole list of scripts, which is probably not so bad since it avoids that the following scripts work on bad data.
You can name a function for all the script 2 like this:
script2.py
def main():
print('Hello World!')
And import script2 function with this:
script1.py
from script2.py import *
main()
I hope this is helpful (tell me if I didn't answer to your question I'm Italian..)
For the life of me i can't figure this one out.
I have 2 applications build in python, so 2 projects in different folders, is there a command to say in the first application like run file2 from documents/project2/test2.py ?
i tried something like os.system('') and exec() but that only seems to work if its in the same folder. How can i give a command a path like documents/project2 and then for example:
exec(documents/project2 python test2.py) ?
short version:
Is there a command that runs python test2.py while that test2 is in a completely different file/project?
thnx for all feedback!
There's a number of approaches to take.
1 - Import the .py
If the path to the other Python script can be made relative to your project, you can simply import the .py. This will cause all the code at the 'root' level of the script to be executed and makes functions as well as type and variable definitions available to the script importing it.
Of course, this only works if you control how and where everything is installed. It's the most preferable solution, but only works in limited situations.
import ..other_package.myscript
2 - Evaluate the code
You can load the contents of the Python file like any other text file and execute the contents. This is considered more of a security risk, but given the interpreted nature of Python in normal use not that much worse than an import under normal circumstances.
Here's how:
with open('/path/to/myscript.py', 'r') as f:
exec(f.read())
Note that, if you need to pass values to code inside the script, or out of it, you probably want to use files in this case.
I'd consider this the least preferable solution, due to it being a bit inflexible and not very secure, but it's definitely very easy to set up.
3 - Call it like any other external program
From a Python script, you can call any other executable, that includes Python itself with another script.
Here's how:
from subprocess import run
run('python path/to/myscript.py')
This is generally the preferable way to go about it. You can use the command line to interface with the script, and capture the output.
You can also pipe in text with stdin= or capture the output from the script with stdout=, using subprocess.Popen directly.
For example, take this script, called quote.py
import sys
text = sys.stdin.read()
print(f'In the words of the poet:\n"{text}"')
This takes any text from standard in and prints them with some extra text, to standard out like any Python script. You could call it like this:
dir | python quote.py
To use it from another Python script:
from subprocess import Popen, PIPE
s_in = b'something to say\nright here\non three lines'
p = Popen(['python', 'quote.py'], stdin=PIPE, stdout=PIPE)
s_out, _ = p.communicate(s_in)
print('Here is what the script produced:\n\n', s_out.decode())
Try this:
exec(open("FilePath").read())
It should work if you got the file path correct.
Mac example:
exec(open("/Users/saudalfaris/Desktop/Test.py").read())
Windows example:
exec(open("C:\Projects\Python\Test.py").read())
Today I managed to run my first Python script ever. I'm a newb, on Windows 7 machine.
When I run python.exe and enter following (Python is installed in C:/Python27)
import os
os.chdir('C:\\Pye\\')
from decoder import *
decode("12345")
I get the desired result in the python command prompt window so the code works fine. Then I tried to output those results to a text file, just so I don't have to copy-paste it all manually in the prompt window. After a bit of Googling (again, I'm kinda guessing what I'm doing here) I came up with this;
I wrote "a.py" script in the C:/Pye directory, and it looked like this;
from decoder import *
decode("12345")
And then I wrote a 01.py file that looked like this;
import subprocess
with open("result.txt", "w+") as output:
subprocess.call(["python", "c:/Pye/a.py"], stdout=output);
I see the result.txt gets created in the directory, but 0 bytes. Same happens if I already make an empty result.txt and execute the 01.py (I use Python Launcher).
Any ideas where am I screwing things up?
You didn't print anything in a.py. Change it to this:
from decoder import *
print(decode("12345"))
In the Python shell, it prints it automatically; but the Python shell is just a helper. In a file, you have to tell it explicitly.
When you run python and enter commands, it prints to standard out (the console by default) because you're using the shell. What is printed in the python shell is just a representation of what object is returned by that line of code. It's not actually equivalent to explicitly calling print.
When you run python with a file argument, it executes that script, line by line, without printing any variables to stdout unless you explicitly call "print()" or write directly to stdout.
Consider changing your script to use the print statement.:
print(decode("12345"))
I just want to have some ideas to know how to do that...
I have a python script that parses log files, the log name I give it as an argument so that when i want to run the script it's like that.. ( python myscript.py LOGNAME )
what I'd like to do is to have two scripts one that contains the functions and another that has only the main function so i don't know how to be able to give the argument when i run it from the second script.
here's my second script's code:
import sys
import os
path = "/myscript.py"
sys.path.append(os.path.abspath(path))
import myscript
mainFunction()
the error i have is:
script, name = argv
valueError: need more than 1 value to unpack
Python (just as most languages) will share parameters across imports and includes.
Meaning that if you do:
python mysecondscript.py heeey that will flow down into myscript.py as well.
So, check your arguments that you pass.
Script one
myscript = __import__('myscript')
myscript.mainfunction()
script two
import sys
def mainfunction():
print sys.argv
And do:
python script_one.py parameter
You should get:
["script_one.py", "parameter"]
You have several ways of doing it.
>>> execfile('filename.py')
Check the following link:
How to execute a file within the python interpreter?